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Alex787 [66]
3 years ago
11

The length of the hypotenuse of a right triangle is 16 inches if the length of one leg is 5 inches what is the approximate lengt

h of the other leg
Mathematics
1 answer:
alekssr [168]3 years ago
8 0
To find the missing side of the triangle, we use pythagorean theorem.
c^2=a^2+b^2
Put values
16^2 = 5^2 +b^2
256=25+b^2
Subtract 25 from both sides.
231=b^2
Take square root.
15.2=b
or
b=15.2 Rounded to nearest tenth.


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AB is a diameter of a circle, center O. C is a point on the circumference of the circle, such that
Solnce55 [7]

Given:

AB is the diameter of a circle.

m∠CAB = 26°

To find:

The measure of m∠CBA.

Solution:

Angle formed in the diameter of a circle is always 90°.

⇒ m∠ACB = 90°

In triangle ACB,

Sum of the angles in the triangle = 180°

m∠CAB + m∠ACB + m∠CBA = 180°

26° + 90° + m∠CBA = 180°

116° + m∠CBA = 180°

Subtract 116° from both sides.

116° + m∠CBA - 116° = 180° - 116°

m∠CBA = 64°

The measure of m∠CBA is 64°.

7 0
3 years ago
Please help out asap thanks
Jet001 [13]

Answer:

there all below

Step-by-step explanation:

Mean: 51.5

There is no mode so you would put "No mode.''

Median: 51

hope this helps <33

6 0
2 years ago
Please help I need it ASAP
Irina-Kira [14]
Y=2x-1





twenty characters
4 0
3 years ago
Solve using Fourier series.
Olin [163]
With 2L=\pi, the Fourier series expansion of f(x) is

\displaystyle f(x)\sim\frac{a_0}2+\sum_{n\ge1}a_n\cos\dfrac{n\pi x}L+\sum_{n\ge1}b_n\sin\dfrac{n\pi x}L
\displaystyle f(x)\sim\frac{a_0}2+\sum_{n\ge1}a_n\cos2nx+\sum_{n\ge1}b_n\sin2nx

where the coefficients are obtained by computing

\displaystyle a_0=\frac1L\int_0^{2L}f(x)\,\mathrm dx
\displaystyle a_0=\frac2\pi\int_0^\pi f(x)\,\mathrm dx

\displaystyle a_n=\frac1L\int_0^{2L}f(x)\cos\dfrac{n\pi x}L\,\mathrm dx
\displaystyle a_n=\frac2\pi\int_0^\pi f(x)\cos2nx\,\mathrm dx

\displaystyle b_n=\frac1L\int_0^{2L}f(x)\sin\dfrac{n\pi x}L\,\mathrm dx
\displaystyle b_n=\frac2\pi\int_0^\pi f(x)\sin2nx\,\mathrm dx

You should end up with

a_0=0
a_n=0
(both due to the fact that f(x) is odd)
b_n=\dfrac1{3n}\left(2-\cos\dfrac{2n\pi}3-\cos\dfrac{4n\pi}3\right)

Now the problem is that this expansion does not match the given one. As a matter of fact, since f(x) is odd, there is no cosine series. So I'm starting to think this question is missing some initial details.

One possibility is that you're actually supposed to use the even extension of f(x), which is to say we're actually considering the function

\varphi(x)=\begin{cases}\frac\pi3&\text{for }|x|\le\frac\pi3\\0&\text{for }\frac\pi3

and enforcing a period of 2L=2\pi. Now, you should find that

\varphi(x)\sim\dfrac2{\sqrt3}\left(\cos x-\dfrac{\cos5x}5+\dfrac{\cos7x}7-\dfrac{\cos11x}{11}+\cdots\right)

The value of the sum can then be verified by choosing x=0, which gives

\varphi(0)=\dfrac\pi3=\dfrac2{\sqrt3}\left(1-\dfrac15+\dfrac17-\dfrac1{11}+\cdots\right)
\implies\dfrac\pi{2\sqrt3}=1-\dfrac15+\dfrac17-\dfrac1{11}+\cdots

as required.
5 0
3 years ago
What is 8 + x ≥ 12 on a number line???
levacccp [35]
8+x ≥ 12
-8 -8
x ≥ 4
(i tried taking the picture)

6 0
3 years ago
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